Consider two observers moving in the x direction with respect to a stationary reference frame

I'll assume that we are using units such that c=1. The two world lines are given by:

x1 = t/3

x2 = t^2/18

Here the "18" must carry the units of time used. So, we'll just assume that some units are chosen and the numbers that come out for times/distances are given in terms of that unit.

The time when the observers meet follows from equating x1 to x2:

x1 = x2 -->

t = 0 and t = 6.

At t = 6, x1 = x2 = 2.

The event where they meet again is thus the event t = 6, x = 2.

Now, you know from elementary special relativity that the relation between proper time tau and coordinate time t is:

tau = t/gamma

where gamma = 1/sqrt[1-v^2]

But you must be very careful about applying equations from special relativity derived for inertial observers to accelerating observers. So, let's apply this equation to observer 1 and then I'll show how you deal with the general case. Observer 1 moves with speed v = 1/3 and therefore:

tau = t/gamma = 6*sqrt[1-1/9] = 4 sqrt[2] = approximately 5.6569

Now let's look at observers that are not necessarily inertial observers. In that case you start from the line element:

ds^2 = dt^2 - dx^2

The space-time interval ds is postulated to be (locally) invariant under general coordinate transformations. Suppose some observer is moving w.r.t. me in some arbitrary way. Let's define two nearby events to be two ticks of a clock that the observer is carrying with him. In my coordinate system where I'm at rest the value for ds is:

ds = sqrt[dt^2 - dx^2] = sqrt[1-(dx/dt)^2] dt

Here dt is the time interval between two ticks of the clock of the observer according to my coordinate system. dx is the distance the clock has travelled in my coordinate system between the two ticks. dx/dt is thus the speed at which the observer is traveling relative to me.

If the observer evaluates the space-time interval between two ticks of the clock then he'll insert for dt the time difference between two clicks as he observes it. for dx he'll take zero, because in his frame the clock is at rest. If we call the time interval in the observer's rest frame d tau, then:

ds = d tau.

Because the coordinate transformation from the original coordinates to the coordinates in which the observer is at rests leaves ds invariant, you have:

ds = d tau = sqrt[1-(dx/dt)^2] dt

This means that the old formula from elementary special relativity is still valid in infinitesimal form. If an observer is moving at constant speed you regain the equation tau = t/gamma, while in the general case you have to integrate the above equation.

Let's integrate the above equation for observer 2. The speed is:

dx/dt = t/9

So, we get:

tau = integral from t=0 to 6 of sqrt[1-(t/9)^2]dt

Substitute in here t = 9 sin(y):

tau = integral from y=0 to p of 9 cos^2 (y)dy

where p = arcsin(2/3)

Using

cos^2(y) = 1/2 + 1/2cos(2y)

allows you to calculate the integral easily:

tau = 9[p/2 + 1/4 sin(2p)]=

9[p/2 + 1/2 sin(p)cos(p)]=

9[p/2 + 1/2 sin(p)sqrt[1- sin^2(p)]] =

9/2 arcsin(2/3) + 3 sqrt[1-4/9] =

9/2 arcsin(2/3) + sqrt[5] =

approximately 5.5198

You see that the two proper times for the two observers don't agree. This illustrates the fact that you can only define invariant space time interval locally for infinitesimally close events. If you only consider inertial observers then the space time interval is a global invariant.

Another example is the well known twin paradox, where one twin returns from space travel and is younger than the other twin. The twin that has traveled and returned must have undergone an acceleration and could not have been an inertial observer.